$\sin{2x} = \frac{2 \tan {x}}{1 + \tan^2{x}}$. Do L.H.S. and R.H.S. have same domain?

Asked Jul 02 '20 at 22:51

Active Jul 02 '20 at 22:51

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I can see that in the function $\sin{2x}$, x can be any real number.

But I can also see that in the function $\frac{2 \tan {x}}{1 + \tan^2{x}}$, x cannot be $n \pi + \frac{\pi}{2}$ where n = any integer.

But both functions are equal. How can two functions which are equal have different domains ?

asked Jul 02 '20 at 22:51

anonymous

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It is understood that the identity applies only for $x$ such that both sides are defined, i.e., $x$ in the intersection of the domains of the two functions. – Brian M. Scott Jul 02 '20 at 22:54
Does this answer your question? – Andrew Chin Jul 02 '20 at 22:57
If $ x=\pi/2+n\pi $, you cannot speak about the second function at all. – hamam_Abdallah Jul 02 '20 at 22:58
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@GeoffreyTrang $$\frac{2\tan x}{1+\tan^2x}=\frac{2\tan x}{\sec^2x}=2\tan x\cos^2x=2\sin x\cos x=\sin2x$$ – Andrew Chin Jul 02 '20 at 22:59
@Andrew Chin I read the answer, but I still cannot understand how L.H.S and R.H.S functions are equal if L.H.S is defined for all real numbers, but R.H.S is not defined for all real numbers, i.e. their domains are unequal. – anonymous Jul 02 '20 at 23:16
@AndrewChin Yes, indeed. – Geoffrey Trang Jul 02 '20 at 23:17
If their domains are unequal, then they are unequal functions. – Andrew Chin Jul 02 '20 at 23:28
@Andrew Chin If $x = \frac{π}{2}$, $\sin{2x} $ is defined but $ \frac{2\tan{x}}{1 + \tan^2{x}}$ is not defined. So, does $\frac{π}{2}$ belongs to the domain of $\sin{2x}$ here ? – anonymous Jul 02 '20 at 23:43
You are correct. – Andrew Chin Jul 02 '20 at 23:45

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